The solution of the equation ${\cos ^2}x - 2\cos x = $ $4\sin x - \sin 2x,$ $\,(0 \le x \le \pi )$ is
$\pi - {\cot ^{ - 1}}\left( {\frac{1}{2}} \right)$
$\pi - {\tan ^{ - 1}}(2)$
$\pi + {\tan ^{ - 1}}\left( { - \frac{1}{2}} \right)$
None of these
Let $S\, = \,\left\{ {\theta \, \in \,[ - \,2\,\pi ,\,\,2\,\pi ]\, :\,2\,{{\cos }^2}\,\theta \, + \,3\,\sin \,\theta \, = \,0} \right\}$. Then the sum of the elements of $S$ is
The smallest positive root of the equation $tanx\, -\, x = 0$ lies on
The solution set of $(5 + 4\cos \theta )(2\cos \theta + 1) = 0$ in the interval $[0,\,\,2\pi ]$ is
Let $S$ be the sum of all solutions (in radians) of the equation $\sin ^{4} \theta+\cos ^{4} \theta-\sin \theta \cos \theta=0$ in $[0,4 \pi]$ Then $\frac{8 \mathrm{~S}}{\pi}$ is equal to ...... .
The number of solutions of $|\cos x|=\sin x$, such that $-4 \pi \leq x \leq 4 \pi$ is.